Rationalize the denominator of $\frac{3}{2\sqrt[3]{5}}$. The answer can be written in the form of $\frac{A\sqrt[3]{B}}{C}$, where $A$, $B$, and $C$ are integers, $C$ is positive, and $B$ is not divisible by the cube of any prime. Find $A+B+C$.
Answer: Since the cube of $5$ is $125$, we multiply the top and bottom by $\sqrt[3]{\frac{125}{5}}$, which is $\sqrt[3]{25}$. $$\frac{3}{2\sqrt[3]{5}} \cdot \frac{\sqrt[3]{25}}{\sqrt[3]{25}}=$$$$\frac{3\sqrt[3]{25}}{10}$$Therefore, $A+B+C=3+25+10=\boxed{38}$.